Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{-5z^2 + 15z + 350}{2z^2 - 40z + 200} \div \dfrac{z + 7}{4z + 12} $
Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{-5z^2 + 15z + 350}{2z^2 - 40z + 200} \times \dfrac{4z + 12}{z + 7} $ First factor out any common factors. $t = \dfrac{-5(z^2 - 3z - 70)}{2(z^2 - 20z + 100)} \times \dfrac{4(z + 3)}{z + 7} $ Then factor the quadratic expressions. $t = \dfrac {-5(z - 10)(z + 7)} {2(z - 10)(z - 10)} \times \dfrac {4(z + 3)} {z + 7} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac { -5(z - 10)(z + 7) \times 4(z + 3)} { 2(z - 10)(z - 10) \times (z + 7)} $ $t = \dfrac {-20(z - 10)(z + 7)(z + 3)} {2(z - 10)(z - 10)(z + 7)} $ Notice that $(z - 10)$ and $(z + 7)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {-20\cancel{(z - 10)}(z + 7)(z + 3)} {2\cancel{(z - 10)}(z - 10)(z + 7)} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $t = \dfrac {-20\cancel{(z - 10)}\cancel{(z + 7)}(z + 3)} {2\cancel{(z - 10)}(z - 10)\cancel{(z + 7)}} $ We are dividing by $z + 7$ , so $z + 7 \neq 0$ Therefore, $z \neq -7$ $t = \dfrac {-20(z + 3)} {2(z - 10)} $ $ t = \dfrac{-10(z + 3)}{z - 10}; z \neq 10; z \neq -7 $